如何使＆QUOT;若＆QUOT;条件的立场在C所有结构项目 [英] How to make "if" condition stand for all struct items in C

问题描述

我想知道如何让如果状态代表在该节点的所有结构项目。它计算只有1项和退出周期。这里是code。即时试图使该检查在数据库中的项目的到期日期的功能。请忽略保加利亚文。

  LIST * sortPr（LIST * pFirst）{    LIST * B = NULL，* P;    LIST * B1，*总线1 = NULL;
    INT yearcur = 1;
    的printf（Vuvedete teku6ta戈迪纳）;
    scanf函数（％d个，＆安培; yearcur）;
    如果（pFirst == NULL）
    {
        的setColor（COL_RED）;
        的printf（Списъкътепразен\\ n）;
        系统（暂停）;
        出口（4）;
    }
    为（P = pFirst; P！= NULL; P = P-＆GT; pNext）
    {
        如果（（yearcur  - 对 - ＆GT; body.year）GT; P-＆GT; body.expd）
        {                B =磷;        }        B1 =为newElement（BUS1，B＆GT;车身）;
        返回B1;
    }
}
 

解决方案

这是因为你有 B1回报您的循环内。这将第一次迭代后终止，并返回 B1 出来的功能。

  LIST * sortPr（LIST * pFirst）{
    LIST * B = NULL，* P;
    LIST * B1，*总线1 = NULL;
    ...
    为（P = pFirst; P！= NULL; P = P-＆GT; pNext）{
        ...        //这将第一次迭代过程本身返回B1
        返回B1;
    }
}
 

I want to know how to make that "if" condition stand for all struct items in the node. It counts just 1 items and quits the cycle. Here is the code. Im trying to make a function that checks the expiry date on a item in a database. Please disregard the bulgarian text.

LIST *sortPr(LIST *pFirst) {

    LIST *b = NULL, *p;

    LIST *b1, *bus1 = NULL;
    int yearcur = 1;
    printf("Vuvedete teku6ta godina");
    scanf("%d", &yearcur);
    if (pFirst == NULL)
    {
        setcolor(COL_RED);
        printf("Списъкът е празен\n");
        system("pause");
        exit(4);
    }
    for (p = pFirst; p != NULL; p = p->pNext)
    {
        if ((yearcur - p->body.year) > p->body.expd)
        {

                b = p;



        }

        b1 = newElement(bus1, b->body);
        return b1;
    }
}

解决方案

This is because you have return b1 inside your for loop. This would terminate it after the first iteration, and return b1 out of the function.

LIST *sortPr(LIST *pFirst) {
    LIST *b = NULL, *p;
    LIST *b1, *bus1 = NULL;
    ...
    for (p = pFirst; p != NULL; p = p->pNext) {
        ...

        // This would return b1 during the first iteration itself
        return b1;
    }
}

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